∡MAD = 30
AD-h⇒∡ADM=90°
De aici rezulta m(∡DMA) = 90°-30° =60°
Astfel,putem aplica teorema de 30°-60°-90°, deci:
Conform T 30°-60°-90°: DM = AM/2 => AM = 10 x 2 = 20
ΔDMA-dr [tex]T.P.\\==>[/tex]
[tex]AD^{2} =20^{2} -10^{2} \\AD^{2} =400-100\\AD=\sqrt{300} \\AD=10\sqrt{3}[/tex]
Fie MF⊥AC⇒ΔMFA si ΔMFC-dr
Avem m(∡DMA)=60°⇒m(∡AMC)=180°-60°=120°
∡AMF=∡CMF=60° (jumatate din masura ∡AMC)
ΔMFA≡ΔMFC (∡MAF=∡MCF=30°) ⇒⇒⇒ΔMAC-isoscel
AM=20=MC
m(∡BAD)=30°
m(∡ADC)=90°⇒m(∡ABD)=60°
AM-mediana
AM imparte BC in doua segmente egale (BM,MC)
BM=MC=20
BC=BM+MC
BC=20+20
BC=40
